How are these pattered in the rule of derivatives?

Find the derivative of dy/dx of the relations:

1.) x^3y+ 2y^4 - x^4 = 2xy

and

2.) x^2 = x+2y/ x - 2y

thanks a lot

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- July 31st 2012, 12:45 AMrcsderivative very tricky : need guide
How are these pattered in the rule of derivatives?

Find the derivative of dy/dx of the relations:

1.) x^3y+ 2y^4 - x^4 = 2xy

and

2.) x^2 = x+2y/ x - 2y

thanks a lot - July 31st 2012, 01:45 AMjgv115Re: derivative very tricky : need guide
If I want to simplify , I obviously can't do it that easily as the variables are wrong.

However, I can apply the chain rule:

Solving for , I get

Evaluate which is . Therefore

Now try apply these principles in your question - July 31st 2012, 02:35 AMrcsRe: derivative very tricky : need guide
ill try sir... ill be back and let you see my answer also... hope i can get it right

thanks - July 31st 2012, 02:40 AMemakarovRe: derivative very tricky : need guide
To be honest, I don't understand this. What is and how is it different from ?

OP, it seems that you need to differentiate an implicit function. Obviously, you need to read how to do this, for example, in Wikipedia. Once you learn the method and read several finished examples, please describe the exact difficulty you are having with this particular problem. The thing is that this forum is not a tutorial service and you can't expect people to explain completely new material to you. The forum is here to help those who know the material but struggle to apply it in a concrete situation. - July 31st 2012, 04:36 AMrcsRe: derivative very tricky : need guide
thank you

- July 31st 2012, 04:37 AMDevenoRe: derivative very tricky : need guide
i'll give you an example, perhaps you'll see how it works:

if x^{2}+y^{2}= 1, find dy/dx.

step 1) differentiate both sides with respect to x:

2x(dx/dx) + 2y(dy/dx) = 0 but dx/dx = 1, so:

2x + 2y(dy/dx) = 0

step 2) solve for dy/dx:

2y(dy/dx) = -2x

dy/dx = -2x/2y = -x/y - July 31st 2012, 05:40 PMx3bnmRe: derivative very tricky : need guide
So this is called implicit differentiation. You can differentiate both sides with respect to where is a function of and :

For the second one:

You can check it to confirm here for the first equation:

differentiation x^3 y +2 y^4 -x^4 - 2xy = 0 - Wolfram|Alpha

And for the second one:

differentiation x^2 - x - 2y/x +2y = 0 - Wolfram|Alpha

Hope this helps. - July 31st 2012, 06:28 PMrcsRe: derivative very tricky : need guide
a million of thanks for the solution sir... the best Helper of this site... --- rcs ----

- July 31st 2012, 07:11 PMx3bnmRe: derivative very tricky : need guide
- September 2nd 2012, 07:32 PMrcsRe: derivative very tricky : need guide
- September 2nd 2012, 09:05 PMProve ItRe: derivative very tricky : need guide
x3bnm read the question AS YOU WROTE IT. If you had used the brackets in the correct place in the first place, you would have been given a correct solution.

Anyway, you now know how to differentiate implicitly, so I suggest you try it out for yourself with the correct question. - September 3rd 2012, 06:05 AMrcsRe: derivative very tricky : need guide
- September 4th 2012, 04:00 PMx3bnmRe: derivative very tricky : need guide
Yes the new solution is given below:

Hope it will help you.

Also you can check the answer at:

differentiation x^2 = (x+2y)/(x-2y) - Wolfram|Alpha - September 7th 2012, 06:50 AMrcsRe: derivative very tricky : need guide
- September 7th 2012, 07:02 AMrcsRe: derivative very tricky : need guide
Such a big help sir. Thanks